Mon Feb Matrix algebra and matrix inverses. Announcements: Warm-up Exercise:

Transcription

1 Math Week 5 notes We will not necessarily finish the material from a given day's notes on that day We may also add or subtract some material as the week progresses, but these notes represent an in-depth outline of what we plan to cover These notes cover material in Mon Feb Matrix algebra and matrix inverses Announcements: Warm-up Exercise: Remember from last week that matrices correspond to linear transformations, and that products of matrices are the matrices of compositions of linear transformations:

2 21 Matrix multiplication Suppose we take a composition of linear transformations: : n m, x = A x, A m n T 2 : m p, T 2 y = B y B p m Then the composition T 2 : n p is linear: i T 2 u v T 2 u v = T 2 u v linear = T 2 u T 2 v T 2 linear T 2 u T 2 v ii T 2 c u T 2 c u = T 2 c u linear = c T 2 u T 2 linear c T 2 u So T 2 : n p is a matrix transformation, by the theorem on the previous page Its j th column is T 2 e j = T 2 A e j = T 2 col j A = B col j A ie the matrix of T 2 : n p is where col j A = a j B a 1 B a 2 B a n B A Summary: For B p m and A m n the matrix product BA p or equivalently And, because BA is the matrix of T 2 n is defined by col j B A = B col j A j = 1 n entry i j B A = row i B col j A i = 1 p, j = 1 n BA x = B A x

3 21 matrix algebrawe already talked about matrix multiplication It interacts with matrix addition in interesting ways We can also add and scalar multiply matrices of the same size, just treating them as oddly-shaped vectors: Matrix algebra: addition and scalar multiplication: Let A m n, B m n be two matrices of the same dimensions (m rows and n columns) Let entry i j A = a i j, entry i j B = b i j (In this case we write A = a i j, B = b i j ) Let c be a scalar Then entry i j A B a i j b i j entry i j c A c a i j In other words, addition and scalar multiplication are defined analogously as for vectors In fact, for these two operations you can just think of matrices as vectors written in a rectangular rather than row or column format Exercise 1) Let A := 3 1 and B := 5 1 Compute 4 A B vector properties of matrix addition and scalar multiplication But other properties you're used to do hold: is commutative A B = B A entry ij A B = a ij b ij = b ij a ij = entry ij B A is associative A B C = A B C the ij entry of each side is a ij b ij c ij scalar multiplication distributes over c A B = ca cb ij entry of LHS is c a ij b ij = c a ij b ij = ij entry of RHS

4 More interesting are how matrix multiplication and addition interact: Check some of the following Let I n be the n n identity matrix, with I n x = x for all x n Let A, B, C have compatible dimensions so that the indicated expressions make sense Then a A B C = AB C (associative property of multiplication) b A B C = A B A C (left distributive law) c A B C = A C B C (right distributive law) d r AB = ra B = A rb for any scalar r e) If A m n then I m A = A and A I n = A Warning: AB BA in general In fact, the sizes won't even match up if you don't use square matrices

5 The transpose operation: Definition: Let B m n = b i j Then the transpose of B, denoted by B T is an n m matrix defined by entry i j B T entry j i B = b j i The effect of this definition is to turn the columns of B into the rows of B T : entry i col j B = b i j entry i row j B T = entry j i B T = b i j And to turn the rows of B into the columns of B T : entry j row i B = b i j entry j col i B T = entry j i B T = b i j Exercise 2) explore these properties with the identity T = Algebra of transpose: a A T T = A b A B T = A T B T c for every scalar r ra T = r A T d (The only surprising property, so we should check it) A B T = B T A T

6 22 matrix inverses We've been talking about matrix algebra: addition, scalar multiplication, multiplication, and how these operations combine But I haven't told you what the algebra on the previous page is good for Today we'll start to find out By way of comparison, think of a scalar linear equation with known numbers a, b, c, d and an unknown number x, a x b = c x d We know how to solve it by collecting terms and doing scalar algebra: a x c x = d b a c x = d b * x = d b a c How would you solve such an equation if A, B, C, D were square matrices, and X was a vector (or matrix)? Well, you could use the matrix algebra properties we've been discussing to get to the step And then if X was a vector you could solve the system * with Gaussian elimination In fact, if X was a matrix, you could solve for each column of X (and do it all at once) with Gaussian elimination But you couldn't proceed as with scalars and do the final step after the * because it is not possible to divide by a matrix Today we'll talk about a potential shortcut for that last step that is an analog of of dividing, in order to solve for X It involves the concept of inverse matrices

7 Matrix inverses: A square matrix A n n is invertible if there is a matrix B n n so that AB = BA = I In this case we call B the inverse of A, and write B = A 1 Remark 1: A matrix A can have at most one inverse, because if we have two candidates B, C with AB = BA = I and also AC = CA = I then BA C = IC = C B AC = BI = B so since the associative property BA C = B AC is true, it must be that B = C Remark 2: In terms of linear transformations, if T : n n is the linear transformation T x = A x, then saying that A has an inverse matrix is the same as saying that T has an inverse linear transformation, : n n with matrix B so that T x = x x n and T y = y y n Your final food for thought question from last Friday explains why linear transformations T : n m only have a chance at having inverse transforms when n = m Exercise 1a) Verify that for A = the inverse matrix is A 1 =

8 Inverse matrices can be useful in solving algebra problems For example Theorem: If A 1 exists then the only solution to Ax = b is x = A 1 b Exercise 1b) Use the theorem and A 1 in 2a, to write down the solution to the system x 2 y = 5 3 x 4 y = 6 Exercise 2a) Use matrix algebra to verify why the Theorem above is true Notice that the correct formula is x = A 1 b and not x = b A 1 (This second product can't even be computed because the dimensions don't match up!)

9 Corollary: If A 1 exists, then the reduced row echelon form of A is the identity matrix proof: For a square matrix, solutions to A x = b always exist and are unique only when A reduces to the identity When A 1 exists, the solutions to A x = b exist and are unique, so rref A must equal the identity Exercise 3 Assuming A is a square matrix with an inverse A 1, and that the matrices in the equation below have dimensions which make for meaningful equation, solve for X in terms of the other matrices: XA C = B

10 But where did that formula for A 1 come from? One Answer: Consider A 1 as an unknown matrix, A 1 = X We want A X = I We can break this matrix equation down by the columns of X In the two by two case we get: A col 1 X col 2 X = In other words, the two columns of the inverse matrix X should satisfy A col 1 X = 1 0, A col 2 X = 0 1 We can solve for both of these mystery columns at once, as we've done before when we had different right hand sides: Exercise 3: Reduce the double augmented matrix to find the two columns of A 1 for the previous example

11 For 2 2 matrices there's also a cool formula for inverse matrices: Theorem: this case, a b c d 1 exists if and only if the determinant D = ad 1 bc of a b c d is non-zero And in a b 1 d b = c d ad bc c a (Notice that the diagonal entries have been swapped, and minus signs have been placed in front of the offdiagonal terms This formula should be memorized) Exercise 4) Check that the magic formula reproduces the answer you got in Exercise 3 for Remark) If ad bc = 0 then rows of A are multiples of each other, so A cannot reduce to the identity, so doesn't have an inverse matrix

12 Exercise 4: Will this always work? Can you find A 1 for A := ?

13 1 5 5 Exercise 5) Will this always work? Try to find B 1 for B := Here's what happens when we try to find the three columns of B 1 : BaugI rref BaugI =

14 Tues Feb Matrix inverses Announcements: Warm-up Exercise:

15 Theorem: Let A n n be a square matrix Then A has an inverse matrix if and only if its reduced row echelon form is the identity In this case the algorithm illustrated in our examples will yield A 1 explanation: By the theorem, we discussed on Monday, when A 1 exists, the linear systems A x = b always have unique solutions (x = A 1 b From our previous discussions about reduced row echelon form, we know that for square matrices, solutions to such linear systems exist and are unique if and only if the reduced row echelon form of A is the identity matrix Thus by logic, whenever A 1 exists, A reduces to the identity In this case that A does reduce to I, we search for A 1 as the solution matrix X to the matrix equation A X = I ie A col 1 X col 2 X col n X = Because A reduces to the identity matrix, we may solve for X column by column as in the examples we've worked, by using a chain of elementary row operations: A I I B, and deduce that the columns of X are exactly the columns of B, ie X = B Thus we know that A B = I To realize that B A = I as well, we would try to solve B Y = I for Y, and hope Y = A But we can actually verify this fact by reordering the columns of I B to read B I and then reversing each of the elementary row operations in the first computation, ie create the reversed chain of elementary row operations, B I I A so B A = I also holds (This is one of those rare times when matrix multiplication actually is commuative) To summarize: If A 1 exists, then solutions x to A x = b always exist and are unique, so the reduced row echelon form of A is the identity If the reduced row echelon form of A is the identity, then A 1 exists, because we can find it using the algorithm above That's exactly what the Theorem claims

16 Saying the same thing in lots of different ways (important because it ties a lot of our Chapter 1-2 ideas together): Can you explain why these are all equivalent? The invertible matrix theorem (page 114) Let A be a square n n matrix Then the following statements are equivalent That is, for a given A, the statements are either all true or all false a) A is an invertible matrix b) The reduced row echelon form of A is the n n identity matrix c) A has n pivot positions d) The equation A x = 0 has only the trivial solution x = 0 e) The columns of A form a linearly independent set f) The linear transformation T x A x is one-one g) The equation A x = b has at least one solution for each b n h) The columns of A span n i) The linear transformation T x A x maps n onto n j) There is an n n matrix C such that C A = I k) There is an n n matrix D such that A D = I l) A T is an invertible matrix

17 Wed Feb Matrix inverses: the product of elementary matrices approach to matrix inverses Announcements: Warm-up Exercise:

18 Exercise 1) Show that if A, B, C are invertible matrices, then A B 1 = B 1 A 1 ABC 1 = C 1 B 1 A 1 Theorem The product of n n invertible matrices is invertible, and the inverse of the product is the product of their inverses in reverse order

19 Saying the same thing in lots of different ways (important because it ties a lot of our Chapter 1-2 ideas together): Can you explain why these are all equivalent? The invertible matrix theorem (page 114) Let A be a square n n matrix Then the following statements are equivalent That is, for a given A, the statements are either all true or all false a) A is an invertible matrix b) The reduced row echelon form of A is the n n identity matrix c) A has n pivot positions

20 d) The equation A x = 0 has only the trivial solution x = 0 e) The columns of A form a linearly independent set f) The linear transformation T x A x is one-one

21 g) The equation A x = b has at least one solution for each b n h) The columns of A span n i) The linear transformation T x A x maps n onto n

22 j) There is an n n matrix C such that C A = I k) There is an n n matrix D such that A D = I l) A T is an invertible matrix

23 Wed Feb Matrix inverses: the product of elementary matrices approach to matrix inverses Announcements: Warm-up Exercise:

24 Exercise 1) Show that if A, B, C are invertible matrices, then A B 1 = B 1 A 1 ABC 1 = C 1 B 1 A 1 Theorem The product of n n invertible matrices is invertible, and the inverse of the product is the product of their inverses in reverse order

25 Our algorithm for finding the inverse of a matrix can be reformulated in terms of a product of so-called "elementary" matrices This product idea will pay off elsewhere To get started, let's notice an analog of the fact that a matrix times a vector is a linear combination of the matrix columns That was in fact how we defined matrix times vector in week 2, although we usually use the dot product way of computing the product entry by entry, instead: Definition (from 14) If A is an m n matrix, with columns a 1, a 2, a n (in m ) and if x n, then A x is defined to be the linear combination of the columns, with weights given by the corresponding entries of x In other words, A x = a 1 a 2 a n x x 1 a 1 x 2 a 2 x n a n Theorem If we multiply a row vector times an n B: proof: We want to check whether m matrix B we get a linear combination of the rows of b 1 x T B = x 1 x 2 x n b 1 : = x 1 b 1 x 2 b 2 x n b n b n where the rows of B are given by the row vectors b 1, b 2, b n This proposed identity is true if and only if its transpose is a true identity But the transpose of the left side is x T B T = B T x T T = B T x = b 1 T b2 T bn T x T T T x 1 b 1 x 2 b 2 x n b n which is the transpose of the right side of the proposed identity So the identity is true QED

26 Exercise 2a Use the Theorem on the previous page and work row by row on so-called "elementary matrix" E 1 on the right of the product below, to show that E 1 A is the result of replacing row 3 A with row 3 A 2 row 1 A, and leaving the other rows unchanged: a 11 a 12 a 13 a 21 a 22 a 23 = a 31 a 32 a 33 2b) The inverse of E 1 must undo the original elementary row operation, so must replace any row 3 A with row 3 A 2 row 1 A So it must be true that Check! E 1 1 = c) What 3 3 matrix E 2 can we multiply times A, in order to multiply row 2 A by 5 and leave the other rows unchanged What is E 2 1? 2d) What 3 3 matrix E 3 can we multiply time A, in order to swap row 1 A with row 3 A? What is E 3 1?

27 Definition An elementary matrix E is one that is obtained by doing a single elementary row operation on the identity matrix Theorem Let E m m be an elementary matrix Let A m n Then the product E A is the result of doing the same elementary row operation to A that was used to construct E from the identity matrix Algorithm for finding A 1 re-interpreted: Suppose a sequence of elementary row operations reduces the n n square matrix A to the identity I n Let the corresponding elementary matrices, in order, be given by Then we have E 1, E 2, E p E p E p 1 E 2 E 1 A = I n E p E p 1 E 2 E 1 A = I n So, A 1 = E p E p 1 E 2 E 1 Notice that E p E p 1 E 2 E 1 = E p E p 1 E 2 E 1 I n so we have obtained A 1 by starting with the identity matrix, and doing the same elementary row operations to it that we did to A, in order to reduce A to the identity I find this explanation of our original algorithm to be somewhat convoluted, but as I said, the matrix product decomposition idea is going to pay dividends elsewhere Also, notice that we have ended up "factoring" A into a product of elementary matrices: A = A 1 1 = 1 Ep E p 1 E 2 E 1 = 1 1 E1 E2 1 Ep 1 E p 1